If f(x) = sin^{-1}left(frac{2 cdot 3^x}{1+9^x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+(3^x)^2}ight)$.Let $3^x = 	an 	heta$,then $	heta = 	an^{-1}(3^x)$.Substituting this into t

Vedclass · Accepted Answer

$\sqrt{3} \log(\sqrt{3})$

Vedclass · Answer

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+(3^x)^2}ight)$.Let $3^x = 	an 	heta$,then $	heta = 	an^{-1}(3^x)$.Substituting this into the function,we get $f(x) = \sin^{-1}\left(\frac{2 	an 	heta}{1+	an^2 	heta}ight) = \sin^{-1}(\sin 2	heta) = 2	heta = 2 	an^{-1}(3^x)$.Differentiating with respect to $x$,we get $f^{\prime}(x) = 2 \cdot \frac{1}{1+(3^x)^2} \cdot \frac{d}{dx}(3^x) = \frac{2 \cdot 3^x \log 3}{1+9^x}$.Evaluating at $x = \frac{1}{2}$,we have $f^{\prime}\left(\frac{1}{2}ight) = \frac{2 \cdot 3^{1/2} \log 3}{1+9^{1/2}} = \frac{2 \sqrt{3} \log 3}{1+3} = \frac{2 \sqrt{3} \log 3}{4} = \frac{\sqrt{3} \log 3}{2} = \sqrt{3} \log(3^{1/2}) = \sqrt{3} \log(\sqrt{3})$.

If $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+9^x}\right)$,then $f^{\prime}\left(\frac{1}{2}\right)$ equals

Similar Questions

The differential coefficient of ${\tan ^{ - 1}}\left( {\frac{x}{{1 + \sqrt {1 - {x^2}} }}} \right)$ with respect to ${\sin ^{ - 1}}x$ is:

Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\tan ^{-1} x$ for $-1 < x < 1$ is:

$\frac{d}{dx} \left\{ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right\} = $

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{1+6x^2} \right) \right) = $ . . . . . .

If $y = \tan^{-1} \left( \frac{5x - x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} \right)$,then $\frac{dy}{dx} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module